Mathematics Higher Tier, Algebraic Fractions
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Question 1
Simplify fully
Answer:
In all of these types of questions you can usually factorise the top and the bottom and then one of
the factors will cancel.
To factorise the top we need to find two numbers that multiply to give 15 and combine to give -8.
These two numbers could be -3 and -5. Therefore the top factorises to give (x -3)(x -5).
To factorise the bottom, it is a bit harder as we have a number greater than 1 in front of the x2 term.
Take the coefficient of the x2 and the units figure (2 and -15) and multiply them together. This gives
-30. Now we must find two numbers that multiply to give -30 but combine to give -7 (the coefficient
of the x term). These two numbers could be +3 and -10. Rewrite the quadratic equation splitting
the x term into these two components:
2x2 – 7x – 15 = 2x2 + 3x – 10x – 15
factorise this into two pairs
x(2x + 3) – 5(2x + 3)
if this is correct then it should be the same factor for both ((2x + 3) in this case)
now factorise again
(2x + 3)(x – 5)
So going back to our original expression we have
and as expected one of the factors (x-5) will cancel on the top and the bottom leaving us with:
3
2 3
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Mathematics Higher Tier, Algebraic Fractions
Question 2
Simplify fully
Answer:
We need to factorise the top and the bottom and hope that something will cancel
To factorise the top we need to first multiply 5 by -3 to get -15. We need to find two numbers that
multiply to get -15 but add to get +14. These two numbers are +15 and -1.
Rewrite the expression splitting the 14x into +15x and -1x
5x2 + 15x – 1x – 3
factorise in pairs
5x(x + 3) – 1(x + 3)
we should have the same thing in both brackets, we do
factorise again
(x + 3)(5x – 1)
To factorise the bottom, we need to first factorise out the 2
2(25x2 – 1)
then recognise that this is the difference of the squares
2(5x – 1)(5x + 1)
putting the fraction back together again in factorised form
we can see that (5x – 1) is on the top and the bottom so can cancel each other out
we are left with
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Mathematics Higher Tier, Algebraic Fractions
Question 3
Solve the equation
=
Answer:
multiply through by (3x – 1)
2–
= multiply through by (2x + 1)
2(2x + 1) – 3(3x – 1) =
multiply through by 5
10(2x + 1) – 15(3x – 1) = 2(3x – 1)(2x + 1)
expand the brackets
10(2x + 1) – 15(3x – 1) = 2(3x – 1)(2x + 1)
20x + 10 – 45x + 15 = 2(6x2 – 1 – 2x + 3x)
group terms
-25x + 25 = 2(6x2 + x – 1)
expand the brackets
-25x + 25 = 12x2 +2x – 2
add 25x to both sides
25 = 12x2 +27x – 2
subtract 25 from boths sides
0 = 12x2 +27x – 27
All the numbers are a multiple of 3 so divide by 3
4x2 +9x – 9 = 0
We can either solve this by factorising or by using the quadratic formula
Factorising
we need to find two numbers that multiply to make -36 (4 x -9) and that add to make 9
these two numbers could be -3 and +12
rewrite the equation splitting the middle term into -3x and 12x
4x2 - 3x + 12x – 9 = 0
factorise in pairs
x(4x - 3) + 3(4x - 3) = 0
factorise again
(4x - 3)(( x + 3) = 0
So either 4x - 3 = 0 or x + 3 = 0
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Mathematics Higher Tier, Algebraic Fractions
If 4x + 3 = 0
4x = 3
x=
If x + 3 = 0
x = -3
quadratic formula
√ a=4
b=9
c = -9
!"! # #!
#
!√$
$
!√$
$
!√
$
!
$
!
$
or !
$
x = or -3
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Mathematics Higher Tier, Algebraic Fractions
Question 4
Simplify fully + %
Answer:
we can first factorise the denominator of the second fraction
x2 + 5x + 6 = (x + 3)(x + 2)
we have
+
In order to add fractions we need a common denominator
let the common denominator be (x + 3)(x + 2)
+
expanding and adding the numerators
&
&
=
=
we can now cancel down the (x + 2) to get
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Mathematics Higher Tier, Algebraic Fractions
Question 5
a) simplify
b) simplify
Answer:
a)
#
=
b) we need a common denominator, (2x – 1)(x + 1)
=
=
= = = www.chattertontuition.co.uk 0775 950 1629
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Mathematics Higher Tier, Algebraic Fractions
Question 6
Solve the equation
+
=1
Answer:
go through and multiply everything by x
4+
=x
multiply everything by x – 2
4(x – 2) + 3x = x(x – 2)
expand the brackets
4x – 8 + 3x = x2 – 2x
group terms
7x – 8 = x2 – 2x
subtract 7x from both sides (so that we move everything on to the side where x2 will be positive)
-8 = x2 – 9x
add 8 to both sides
0 = x2 – 9x + 8
x2 – 9x + 8 = 0
we can factorise as we can see two numbers that multiply to make 8 and add to make -9 (-1 and -8)
(x – 1)(x – 8) = 0
if two things multiply to give 0 then either x -1 = 0 or x – 8 = 0
If x – 1 = 0 then x = 1
or x – 8 = 0 so x = 8
x = 1 or 8
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Mathematics Higher Tier, Algebraic Fractions
Question 7
Simplify fully
Answer:
we can cancel the 2 with the 10 to give
(x + 1)2
5(x + 1)
we can cancel (x + 1) as well
1
5
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Mathematics Higher Tier, Algebraic Fractions
Question 8
Solve
+
=
1
Answer:
first multiply both sides by (x + 3)
4+
= (x + 3)
multiply both sides by (2x – 1)
4(2x – 1) + 3(x + 3) = (x + 3)(2x – 1)
expand
8x – 4 + 3x + 9 = 2x2 + 6x – x – 3
group terms
11x + 5 = 2x2 + 5x – 3
subtract 11x from both sides
5 = 2x2 – 6x – 3
subtract 5 from both sides
0 = 2x2 – 6x – 8
rewrite
2x2 – 6x – 8 = 0
divide by 2
x2 – 3x – 4 = 0
factorise
(x – 4)(x + 1) = 0
so if two brackets multiply to give 0 then one or both of the brackets must equal 0
x = 4 or x = -1
)
)
)
)
to check: put back x = 4 into original equation to get + = = 1
and put x = -1 into original equation to get + = 2 + -1 = 2 – 1 = 1
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Mathematics Higher Tier, Algebraic Fractions
Question 9
Simplify fully
% Answer:
We need to factorise the top and the bottom and hopefully there will be a common factor which will
cancel
Factorise the top
We need to first multiply 6 by -1 to get -6. We need to find two numbers that multiply to give -6 and
combine to give 1. These two numbers are +3 and -2. Rewrite the expression splitting the x term
into +3x and -2x.
6x2 + 3x – 2x – 1
factorise in pairs
3x(2x + 1) – (2x + 1)
we should have the same thing in both brackets which we do
factorise again
(2x + 1)(3x – 1)
Factorise the bottom
this is the difference of two squares
(2x + 1)(2x – 1)
so we have
=
=
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Mathematics Higher Tier, Algebraic Fractions
Question 10
Solve
+
=
Answer:
in order to add fractions we need a common denominator
6 is a multiple of 2 and 3 so use 6 as the denominator
=
&
=
&
We have
&
&
+
&
=
=
multiply both sides by 6
* 5x = = divide both sides by 5
x = = 1.5
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Mathematics Higher Tier, Algebraic Fractions
Question 11
Simplify fully
%
1 + Answer:
In order to add fractions we need a common denominator, but before we do that we can probably
simplify the fraction. We need to factorise the top and usually something will cancel with the
bottom.
find two numbers that multiply to give -6 but add to give 1: the two numbers are +3 and -2
x2 + x – 6 = (x + 3)(x – 2)
so we have 1 + now we can see that the factor (x – 2) will cancel with the top and the bottom to give
1+
Using a common denominator of (x + 4) we can put 1 as
+ =
)
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Mathematics Higher Tier, Algebraic Fractions
Question 12
Solve the equation
–
=3
Answer:
in order to subtract (or add) fractions you must have a common denominator (number on the
bottom)
the common denominator here could be 6
multiply the first fraction by 3 on the top and the same on the bottom
multiply the second fraction y 2 on the top and the same on the bottom
&
–
=
&
3
=
&
3
multiply both sides by 6
3(x + 3) – 2(x – 2) = 18
expand the brackets
3x + 9 – 2x + 4 = 18
group terms
x + 13 = 18
subtract 13 from both sides
x=5
to check put this value of x back into the original equation
$ – = – = 4 – 1 = 3
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Mathematics Higher Tier, Algebraic Fractions
Question 13
a) Simplify fully
b) Simplify fully
Answer:
a) the term (x + 10) cancels from the top and the bottom 4 times leaving just x + 10
another way of seeing this is:
*
*
*
*
*
=
*
*
*
*
x + 10
b) for this type of question we need to factorise the top and the bottom and then usually one of the
factors will cancel
x2 – 25 this is the difference of the square and factorises to give (x – 5)(x + 5)
x2 + 7x + 10 we need to find two numbers that multiply to give +10 and add to give +7. The two
numbers are 2 and 5. So we have (x + 2)(x + 5)
so our fraction becomes (x +5) cancels to give = www.chattertontuition.co.uk 0775 950 1629
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Mathematics Higher Tier, Algebraic Fractions
Question 14
Solve
– =
2
Answer:
multiply both sides by 2
(x + 1) – = 4
multiply both sides by 5
5(x + 1) – 2(x – 3) = 20
expand the brackets
5x + 5 – 2x + 6 = 20
group terms
3x + 11 = 20
subtract 11 from both sides
3x = 9
divide both sides by 3
x=3
remember to check your answer:
+ =
2, 2 + 0 = 2
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Mathematics Higher Tier, Algebraic Fractions
Question 15
Solve
+
=
2
Answer:
to get rid of the fractions go through and multiply by 4 (and then by the 3)
multiply by 4
(2x – 1) + = 8
multiply by 3
3(2x – 1) + 4(x + 2) = 24
expand brackets
6x – 3 + 4x + 8 = 24
group terms
10x + 5 = 24
subtract 5 from both sides
10x = 19
divide both sides by 10
x = 1.9
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Mathematics Higher Tier, Algebraic Fractions
Question 16
Show that the following identity hold for all values of x
– %
+ Answer:
The + sign means that we have an identity rather than an equation. An identity means that the LHS
equals the RHS for all values of x. Whereas an equation means that there is only one (or a finite
number of solutions that x could be). For example 6x + 3x + 9x is an identity because it works for all
x but 6x + 3x = 9 is an equation because it only works for one value of x.
In order to add or subtract fractions you need a common denominator. Let the common
denominator be x(x + 2) (which is just the product of the two denominators)
=
and
=
Now we can subtract the two fractions to get
, working on just the numerator we have
(x – 3)(x + 2) – x(x – 2) = x2 – 3x + 2x – 6 – x2 + 2x
grouping terms
x-6
putting back together again with the denominator we have
6
2
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Mathematics Higher Tier, Algebraic Fractions
Question 17
Simplify fully
+ %
Answer:
We need a common denominator (number on the bottom)
Convert both so that they have a common denominator of 12
&
= !
= now we can add them to get
! +
=
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Mathematics Higher Tier, Algebraic Fractions
Question 18
Solve the equation
+
=
4
Answer:
go through and multiply both sides by 5 to get
( x + 4) + 5(x -2) = 20
3
multiply both sides by 3
3(x + 4) + 5(x – 2) = 60
expand
3x + 12 + 5x – 10 = 60
group terms
8x + 2 = 60
subtract 2 from both sides
8x = 58
divide both sides by 8
x = 7.25
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Mathematics Higher Tier, Algebraic Fractions
Question 19
Simplify fully
Answer:
for these type of questions we need to factorise both the top and the bottom and then something is
bound to cancel down
x2 – 9 = (x + 3)(x – 3) by difference of the square
x2 + 3x = x(x + 3) as x goes into both terms
so we have
=
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Mathematics Higher Tier, Algebraic Fractions
Question 20
Simplify fully
%
Answer:
for this type of question you usually need to factorise the top and the bottom and they are bound to
have one factor in common which you can cancel
working on the numerator:
to factorise 12x2 – 36x + 15 we can first see that 3 goes into all terms so we can first take that out
3(4x2 – 12x + 5)
Then work on factorising 4x2 – 12x + 5
As we have a number in front of the x2, it makes the factorisation a bit harder. We must first
multiply 4 and 5 together to get 20 then we need to find two numbers that multiply together to give
20 and add together to give -12. This would be -2 and -10.
Rewrite 4x2 – 12x + 5 as 4x2 –2x – 10x + 5 (this is just splitting -12x into the two components -2x and 10x)
Now factorise in pairs
2x(2x – 1) – 5(2x – 1)
We should have the same in both brackets (which we do)
As (2x – 1) goes into both terms we can factorise that out to get
(2x – 1)(2x – 5)
Remember we had already factorised the 3 out so the numerator should be
3(2x – 1)(2x – 5)
Working on the denominator:
First we can factorise 3 out to give 3(4x2 – 1)
4x2 – 1 is the difference of two squares so can be expressed as (2x – 1)(2x + 1)
We have
3(2x – 1)(2x + 1)
Putting the numerator and denominator back together
3(2x – 1)(2x – 5) =
3(2x – 1)(2x + 1)
3(2x – 1)(2x – 5)
3(2x – 1)(2x + 1)
We can see that the 3s will cancel out and so will (2x – 1) so we are left with
2 5
2 1
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Mathematics Higher Tier, Algebraic Fractions
Question 21
Simplify fully
%
%
Answer:
factorise the top and the bottom
x2 + 6x = x(x + 6)
x2 – 36 = (x + 6)(x – 6) difference of the square
so we have
&
&
– &
=
– &
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Mathematics Higher Tier, Algebraic Fractions
Question 22
Solve
=
1
Answer:
first multiply by x
2 - = x
now multiply by (x + 1)
2(x + 1) – x = x(x + 1)
expand the brackets
2x + 2 – x = x2 + x
group terms
x + 2 = x2 + x
subtract x from both sides
2 = x2
x2 = 2
square root both sides
x = 1.414
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